I want to calculate the work done by friction if the length $L$ of uniform rope on the table slides off. There is friction between the cord and the table with coefficient of kinetic friction $\mu_k$.

$$ W = \int F \cdot d \vec{s}$$ I think it would be: $$ W_{fr} = \frac M L g \int_{0}^{L} dx$$

But the solutions (which could be mistaken) say: $$ dW_{fr} = \mu_k \frac M L g \, x \, dx$$ which is then integrated.

Should there be an $x$ in the integral? I don't think there should be because you are summing up over an infinitesimal displacement $dx$ and the force of friction is not proportional to the displacement at any instant (I think).

enter image description here

  • $\begingroup$ Do you know that your expression is dimensionally incorrect? $\endgroup$
    – ABC
    Commented Apr 6, 2013 at 3:27

1 Answer 1


Let $x$ denote the length of the rope that is on the table, then $$ m(x) = \frac{M}{L}x $$ is the mass of the rope on the table. It follows that the force of friction on the rope on the table is $$ f(x) = \mu_k m(x) g = \mu_k\frac{M}{L}xg $$ if the rope moves an amount $dx$ then the work done by friction is $$ dW = f(x)dx = \mu_k\frac{M}{L}gx dx $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.